Of course there is a method!
The method is to transform the coordinates in a way that the metric becomes conformally flat in the new coordinates. Then you transform the coordinates somehow that the new coordinates in the flat part of the metric have a finite range. Then you simply draw the flat part of the metric in the new coordinates.

The method is to transform the coordinates in a way that the metric becomes conformally flat in the new coordinates. Then you transform the coordinates somehow that the new coordinates in the flat part of the metric have a finite range. Then you simply draw the flat part of the metric in the new coordinates.

I don't think this is quite right. Conformal flatness is an intrinsic property of a spacetime, so a coordinate transformation doesn't change this property.

Conformal flatness is a hypothesis that must hold for the original spacetime (or it has to be spherically symmetric).

I'm not sure that there is any totally automatic process that works, but if I had to sketch one, it might be something like this:

1. Check that your spacetime has the right properties to make a Penrose diagram possible. This is basically either conformal flatness or rotational symmetry.

2. Reduce the dimensionality to 1+1, e.g., if there's spherical symmetry, reduce every 2-sphere of symmetry to a single point.

3. Try to find a single coordinate patch that covers the whole spacetime. You want this because any Penrose diagram is going to end up on a piece of paper where the paper's (x,y) coordinates cover the whole thing. But I don't think this is always possible, nor is it always trivial if it is possible (e.g., historically, people didn't realize that the event horizon of the Schwarzschild spacetime was really just a coordinate singularity). In some cases this may not be possible, but you may be able to get away with doing something like representing a torus on the paper with dashed lines indicating two lines to be identified.

4. Find a conformal transformation that shrinks your spacetime down so that it fits on the page. A conformal transformation can be represented by a change of metric from the original metric ##g## to an unphysical metric ##\Omega^2 g##, where ##\Omega## depends on the coordinates. Essentially you make ##\Omega## blow up in distant regions so that the unphysical space is finite in size.

I don't think this is quite right. Conformal flatness is an intrinsic property of a spacetime, so a coordinate transformation doesn't change this property.

Conformal flatness is a hypothesis that must hold for the original spacetime (or it has to be spherically symmetric).

Yeah, I should have been more careful. Actually only one part of the metric becomes conformally flat. For example when you write the Schwarzschild metric in Kruskal coordinates, you have ## ds^2=\frac{4r_s^3}{r}e^{-r/r_s}(dT^2-dR^2)-r^2 d\Omega^2 ##. As you can see, the ## \Omega=const ## part of the metric is conformally flat and that's the part of the metric we'll use for the Penrose diagram.

This is something you can find in quite literally every standard textbook on GR. You should learn Penrose diagrams properly from an actual textbook instead of trying to gleam a good understanding of it from a thread. For example see chapters 6 and 12 of Wald.

Yeah, I should have been more careful. Actually only one part of the metric becomes conformally flat. For example when you write the Schwarzschild metric in Kruskal coordinates, you have ## ds^2=\frac{4r_s^3}{r}e^{-r/r_s}(dT^2-dR^2)-r^2 d\Omega^2 ##. As you can see, the ## \Omega=const ## part of the metric is conformally flat and that's the part of the metric we'll use for the Penrose diagram.

Hmm...well, any 2-dimensional space is conformally flat, so I don't think it's necessary to display the 2-dimensional metric explicitly, in a particular coordinate system, in order to show that it's conformally flat.

The Rindler coordinates are just a different set of coordinates for describing Minkowski space. The Penrose diagram for a particular spacetime is independent of what coordinates you use to describe it. Therefore the Penrose diagram is going to be the same as the standard Penrose diagram for Minkowski space.

Hmm...well, any 2-dimensional space is conformally flat, so I don't think it's necessary to display the 2-dimensional metric explicitly, in a particular coordinate system, in order to show that it's conformally flat.

But e.g. the ## \Omega=const## part of the Schwarzschild metric (##ds^2=(1-\frac{r_s}{r})dt^2-(1-\frac{r_s}{r})^{-1} dr^2 ##) is not manifestly conformally flat.

What does it mean, that a metric is "conformally flat"? somewhere in my notes I have the metric about a schwarzschild singularity where the light cones everywhere run off at 45 degrees in transformed (r,ct) Would this be it?

What does it mean, that a metric is "conformally flat"? somewhere in my notes I have the metric about a schwarzschild singularity where the light cones everywhere run off at 45 degrees in transformed (r,ct) Would this be it?

One way of defining it is that a spacetime is conformally flat if, for any point, there is a neighborhood around that point in which coordinates exist such that the metric takes the form AB, where A is a scalar field and B is the standard form of the Minkowski metric.

I am sorry but i do not understand your answer.
Let us take the example if find in thread 5
The change of coordinates is
## x= 1/a e^{a\xi} cosh (a\eta)##
## t= 1/a e^{a\xi} sinh (a\eta)##
so we have ##ds^2 = - dt^2 + dx^2 = e^{2a\xi} (- d\eta^2 + d\xi^2)##
here x is positive.
I have the a conformally flat metric on the right Rindler wedge ## (\eta , \xi)##
What is the new change of coordinates which will map this plane to the square ##\mathcal{R} ## (look at fig 9)?